Does entropy depend on reversibility?

1. When going through a cycle of a process $$A \rightarrow B \rightarrow A$$, is the change in the entropy of the system always equal to $$0$$? Does the reversibility of the process change anything (done reversibly vs irreversibly)?
2. When going through a cycle of a process $$A \rightarrow B \rightarrow A$$, is the change in the entropy of the surroundings always equal to $$0$$? Again, does reversibility have any effect on this?

How does the Clausius Inequality relate to these?

1.) When going through a cycle of a process (A-->B-->A), is the change in the entropy of the system always equal to 0?

Yes. Entropy is a state property. For any cycle that returns the system to its original state, all state properties such as entropy, internal energy, enthalpy, etc. of the system return to their original values.

Does the reversibility of the process change anything (done reversibly vs irreversibly)?

Yes. For a reversible cycle (cycle consisting of all reversible processes) the change in entropy of both the system and the surroundings is zero. For an irreversible cycle (cycle consisting of at least one irreversible process), the change in entropy of the system is zero, but the change in entropy of the surroundings is greater than zero.

For a reversible process, the increase/decrease in entropy of the system equals the decrease/increase in entropy of the surroundings, so that the change in entropy of the combination of the system and surroundings is zero. An example is a reversible isothermal (constant temperature) expansion of an ideal gas. During the process the entropy of the gas increases by $$+\frac{Q}{T}$$ where $$Q$$ is the heat transferred to the gas from the surroundings and $$T$$ is the temperature of the gas. For the process to be reversible the temperature of the gas has to be in equilibrium with the temperature of the surroundings (equal temperatures). The change in entropy of the surroundings is therefore $$-\frac{Q}{T}$$. The total or combination entropy change (system + surroundings) is then zero.

2.) When going through a cycle of a process (A-->B-->A), is the change in the entropy of the surroundings always equal to 0? Again, does reversibility have any effect on this.

As indicated in the answer to (1) for a complete cycle the change in entropy of the surroundings is zero only if the system cycle is reversible. It is greater than zero if the system cycle is irreversible.

Hope this helps.

• @Jeffrey J Weimer I see that you edited my answer and that it indicates I approved the edit. But I don't recall doing so. May have been inadvertent since I was reviewing comments on my mobile device and may have inadvertently pressed some key. In any case while I don't disagree with your edit, it really wasn't necessary. My wording applied to a reversible CYCLE. And for that, the change in entropy of both the system and surroundings is in fact zero. The OP started talking about a cycle then switched to process. Commented Mar 2, 2020 at 3:17
• Your qualification about it being the combination would be needed if I were only talking about a reversible PROCESS. For that, an increase/decrease in entropy of the system would equal the decrease/increase in entropy of the surroundings, i.e., the combination is zero. I am going to edit my answer to point out the difference between a reversible process and a reversible cycle. Just wanted you to see this first to understand why. Thanks Commented Mar 2, 2020 at 3:17
• The entropy change of the system for an isothermal reversible path is not zero. The qualification is needed. Commented Mar 2, 2020 at 15:43
• @JeffreyJWeimer Did you see my latest revision. I didn't say the entropy change for an isothermal reversible path is zero. I said for an expansion there is a positive change in entropy equal to the negative change in entropy of the surroundings. For a reversible cycle the entropy change of both the system and surroundings is zero. Commented Mar 2, 2020 at 15:51
• Your latest revision corrects the ambiguously and perhaps even incorrectly phrased statement in your initial post. It validates my reason for the necessity of my edit. Commented Mar 2, 2020 at 16:43

The change in entropy of a reversible system $$\Delta S = \int \frac{d Q_{rev}}{T}$$ is zero over a closed path, say $$A \rightarrow B \rightarrow A$$, as per the Clausius theorem. Similarly, the change in the entropy of the reservoir or surroundings is also vanishing in the case of reversible processes, although that is not the case for irreversible processes.